A LOGARITHMIC FOURTH-ORDER PARABOLIC ... - Project Euclid

c 2006 International Press °

COMM. MATH. SCI. Vol. 4, No. 2, pp. 275–290

A LOGARITHMIC FOURTH-ORDER PARABOLIC EQUATION AND RELATED LOGARITHMIC SOBOLEV INEQUALITIES∗ § ¨ JEAN DOLBEAULT† , IVAN GENTIL‡ , AND ANSGAR JUNGEL

Abstract. A logarithmic fourth-order parabolic equation in one space dimension with periodic boundary conditions is studied. This equation arises in the context of fluctuations of a stationary nonequilibrium interface and in the modeling of quantum semiconductor devices. The existence of global-in-time non-negative weak solutions and some regularity results are shown. Furthermore, we prove that the solution converges exponentially fast to its mean value in the “entropy norm” and in the Fisher information, using a new optimal logarithmic Sobolev inequality for higher derivatives. In particular, the rate is independent of the solution and the constant depends only on the initial value of the entropy. Key words. Cauchy problem, higher-order parabolic equations, existence of global-in-time solutions, long-time behavior, Fisher information, entropy–entropy production method, logarithmic Sobolev inequality, Poincar´ e inequality. AMS subject classifications.

35K35, 35K55, 35B40

1. Introduction This paper is concerned with the study of some properties of weak solutions to a nonlinear fourth-order equation with periodic boundary conditions and related logarithmic Sobolev inequalities. More precisely, we consider the problem ut + (u(logu)xx )xx = 0,

u(·,0) = u0 ≥ 0

in S 1 ,

(1.1)

where S 1 is the one-dimensional torus parametrized by a variable x ∈ [0,L] for some fixed L > 0. Recently equation (1.1) has attracted the interest of many mathematicians since it possesses some remarkable properties. For instance, it is a one-homogeneous equation which is a simple example of a generalization of the heat equation to higher-order operators. The solutions are non-negative and there are several Lyapunov functionals. A formal calculation shows that the entropy is non-increasing: Z Z d 2 u(logu − 1)dx + u|(logu)xx | dx = 0. (1.2) dt S 1 1 S R Another example of a Lyapunov functional is S 1 (u − logu)dx which formally yields Z Z d 2 (u − logu)dx + |(logu)xx | dx = 0. (1.3) dt S 1 1 S ∗ Received: May 27, 2005; accepted (in revised version): February 10, 2006. Communicated by Anton Arnold. The authors acknowledge partial support from the Project “Hyperbolic and Kinetic Equations” of the European Union, grant HPRN-CT-2002-00282, and from the DAAD-Procope Program. The last author has been supported by the Deutsche Forschungsgemeinschaft, grants JU359/3 (Gerhard-Hess Award) and JU359/5 (Priority Program “Multi-scale Problems”). † Ceremade (UMR CNRS 7534), Universit´ e Paris Dauphine, Place de la Lattre de Tassigny, 75775 Paris, C´ edex 16, France ([email protected]). ‡ Ceremade (UMR CNRS 7534), Universit´ e Paris Dauphine, Place de la Lattre de Tassigny, 75775 Paris, C´ edex 16, France ( [email protected]). § Institut f¨ ur Mathematik, Universit¨ at Mainz, Staudingerweg 9, 55099 Mainz, Germany ([email protected]).

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A LOGARITHMIC FOURTH-ORDER PARABOLIC EQUATION

The last property is used to prove that solutions of (1.1) are non-negative. Indeed, a Poincaré inequality shows that logu is bounded in H 2 (S 1 ) and hence in L∞ (S 1 ), which implies that u ≥ 0 in S 1 × (0,∞). We prove this result rigorously in section 2. Notice that the equation is of higher order and no maximum principle argument can be employed. More Lyapunov functionals of (1.1) have been found in [4, 5]; for a systematic study we refer to [12]. Equation (1.1) has been first derived in the context of fluctuations of a stationary non-equilibrium interface [8]. It also appears as a zero-temperature zero-field approximation of the quantum drift-diffusion model for semiconductors [1] which can be derived by a quantum moment method from a Wigner-BGK equation [7]. The first analytical result has been presented in [4]; there the existence of local-in-time classical solutions with periodic boundary conditions has been proved. A global-in-time existence result with homogeneous Dirichlet-Neumann boundary conditions has been obtained in [13]. However, up to now, no global-in-time existence result is available for the problem (1.1). Our proof is an adaption of the method of [13]; we present the complete proof since we need the approximation scheme for the subsequent sections. The long-time behavior of solutions has been studied in [5] using periodic boundary conditions under restrictive regularity conditions on the initial data, in [15] with homogeneous Dirichlet-Neumann boundary conditions and finally, in [11] employing non-homogeneous Dirichlet-Neumann boundary conditions. In particular, it has been shown that the solutions converge exponentially fast to their steady state in various norms and, see [13], in terms of the entropy. The decay rate has been numerically computed in [6]. We also mention the work [14] in which a positivity-preserving numerical scheme for the quantum drift-diffusion model has been proposed. Concerning the multi-dimensional problem, there exists only the work [10] in which the existence of global-in-time weak solutions has been proved. In the last years the question of non-negative or positive solutions of fourthorder parabolic equations has also been investigated in the context of lubrication-type equations, like the thin film equation ut + (f (u)uxxx )x = 0 (see, e.g., [2, 3]), where typically, f (u) = uα for some α > 0. This equation is of degenerate type which makes the analysis easier than for (1.1), at least concerning the positivity property. Notice that (1.1) is not of degenerate type. In this paper we show the following results. First, the existence of global-intime weak solutions is shown under a rather weak conditionR on the initial datum u0 . We only assume that u0 ≥ 0 is measurable and such that S 1 (u0 − logu0 )dx < ∞. Compared to [4], we do not impose any smallness condition on u0 . We are able to prove that the solution is non-negative. The existence proof is based on a semidiscrete formulation of (1.1). The semi-discrete problem has a strictly positive and smooth solution. This property enables us to prove the long-time behavior of solutions rigorously. √ Our second result is concerned with regularity issues. We prove that, if u0 ∈ H 1 (S 1 ), √ u ∈ L∞ (0,T ;H 1 (S 1 )) ∩ L2 (0,T ;H 3 (S 1 )) for all T > 0. Although one might obtain more regularity results from this (see Remark 3.4), we are interested in applying R this √ property in order to show an exponential decay rate of the Fisher information S 1 |( u)x (·,t)|2 dx.

JEAN DOLBEAULT, IVAN GENTIL AND ANSGAR JUNGEL

277

The third and main result of this paper is the exponential time decay of the solutions, i.e., we show that the solution constructed in Theorem 2.1 converges expoR nentially fast to its mean value u ¯ = u(x,t)dx/L: for all t > 0, µ ¶ Z Z ³u ´ u(x,t) 0 dx ≤ e−M1 t u0 log u(x,t)log dx, (1.4) u ¯ u ¯ S1 S1 Z Z √ √ |( u0 )x |2 dx, (1.5) |( u)x (·,t)|2 dx ≤ e−M2 t S1

S1

where M1 = 32π 4 /L4 , M2 = 16µπ 4 /L4 , and µ = 1.646169... The constant M1 is easily obtained by linearization in the asymptotic regime. It also shows up in [5] (as the decay rate in a different norm) but only in the more restrictive H 1 setting. In [5], an exponential convergence rate in the Lp norm has been given. Then, in principle, the decay rate in the “entropy norm” (1.4) could be derived by letting p → 1. However, the decay rate of [5] contains the factor p − 1 which vanishes in the limit such that no decay rate in the “entropy norm” can be deduced. In [15], the exponential decay of the relative entropy is established, but with a rate which depends on the initial data. Here, M1 and M2 are independent of the solution and the constant on the right-hand side of (1.4) is optimal; it is simply the initial value of the relative entropy. Thus, both decay results (1.4) and (1.5) are new. Our proof is based on the entropy–entropy production method. For the proof R 2 of (1.4), we show that the entropy production term u|(logu)xx | dx in (1.2) can be bounded from below by the entropy itself yielding Z Z ³u´ ³u´ d ulog dx + M1 ulog dx ≤ 0. dt S 1 u ¯ u ¯ S1 Then Gronwall’s inequality gives (1.4). For the proof of (1.5) we first need to show that the following entropy–entropy production inequality holds for some µ > 0, Z Z √ √ d |( u)xxx |2 dx ≤ 0, |( u)x (·,t)|2 dx + µ dt S 1 S1 and then we apply the Poincaré inequality and Gronwall’s lemma. The proof of the above inequality is based on the algorithmic entropy construction method recently developed in [12]. The lower bound for the entropy production in (1.2) is obtained through a logarithmic Sobolev inequality in S 1 . We show (see Theorem 4.1) that any function u ∈ H n (S 1 ) (n ∈ N) satisfies ! Ã µ ¶2n Z ¯ Z ¯ 2 u L ¯ (n) ¯2 2 u log dx ≤ 2 (1.6) ¯u ¯ dx, kuk2L2 (S 1 ) 2π S1 S1 R where kuk2L2 (S 1 ) = u2 dx/L, and the constant is optimal. As already mentioned in the case n = 2, the proof of this result uses the entropy–entropy production method. Entropy estimates are interesting for the following reason. The L1 norm of a solution u to (1.1) is preserved by the evolution. It is therefore natural to look for a convergence of u to its average u ¯ measured in L1 rather than in Lp , p > 1. As noted in many papers, the limit of such Lp estimates as p → 1, p > 1, is the entropy rather than the L1 norm itself. The convergence of u to u ¯ is then a consequence of the

278


standard Csiszár-Kullback inequality. Exactly as for the heat equation, p = 1 looks as a threshold from the point of view of the existence theory and for the optimality of the estimates on the asymptotic behaviour. This work is a step towards a deeper understanding of both entropy methods and higher-order equations. The paper is organized as follows. In section 2 the existence of weak solutions is proved. Section 3 is concerned with the regularity result. Then, section 4 is devoted to the proof of the optimal logarithmic Sobolev inequality (1.6). Finally, in section 5, the exponential time decay (1.4) and (1.5) are shown. 2. Existence of solutions 1 RTheorem 2.1. Let u0 : S → R be a nonnegative measurable function such that (u0 − logu0 )dx < ∞ and let T > 0. Then there exists a global weak solution u of S1 (1.1) satisfying 1,10/9

u ∈ L5/2 (0,T ;W 1,1 (S 1 )) ∩ Wloc (0,T ;H −2 (S 1 )), u ≥ 0 in S 1 × (0,∞), logu ∈ L2 (0,T ;H 2 (S 1 )), and for all T > 0 and all smooth test functions φ, Z T Z TZ hut ,φiH −2 ,H 2 dt + u(logu)xx φxx dxdt = 0. 0

S1

0

The initial datum is satisfied in the sense of H −2 (S 1 ) := (H 2 (S 1 ))∗ . In order to prove this theorem, we first transform (1.1) by introducing the new variable u = ey as in [13]. Then (1.1) becomes (ey )t + (ey yxx )xx = 0,

y(·,0) = y0

in S 1 ,

(2.1)

where y0 = logu0 . In order to prove the existence of solutions to this equation, we semi-discretize (2.1) in time. For this, let T > 0, and let 0 = t0 < t1 < ··· < tN = T with tRk = kτ be a partition of [0,T let yk−1 ∈ H 2 (S 1 ) with R R ] with τ = T /N . Furthermore, R exp(yk−1 )dx = u0 dx and (exp(yk−1 ) − yk−1 )dx ≤ (u0 − logu0 )dx given. Then we solve recursively the elliptic equations 1 yk (e − eyk−1 ) + (eyk (yk )xx )xx = 0 τ

in S 1 .

(2.2)

Lemma 2.2. There exists a solution yk ∈ C ∞ (S 1 ) of (2.2). Proof. Set z = yk−1 . We consider first for given ε > 0 the equation 1 (ey yxx )xx − εyxx + εy = (ez − ey ) in S 1 . τ

(2.3)

In order to prove the existence of a solution to this approximate problem we employ the Leray-Schauder theorem. For this, let w ∈ H 1 (S 1 ) and σ ∈ [0,1] be given, and consider a(y,φ) = F (φ)

for all φ ∈ H 2 (S 1 ),

where for all y,φ ∈ H 2 (S 1 ), Z a(y,φ) = (ew yxx φxx + εyx φx + εyφ)dx, S1

F (φ) =

σ τ

(2.4) Z (ez − ew )φdx. S1


279

Clearly, a(·,·) is bilinear, continuous and coercive on H 2 (S 1 ) and F is linear and continuous on H 2 (S 1 ). (Here we need the additional ε-terms.) Therefore, the LaxMilgram lemma provides the existence of a solution y ∈ H 2 (S 1 ) of (2.4). This defines a fixed-point operator S : H 1 (S 1 ) × [0,1] → H 1 (S 1 ), (w,σ) 7→ y. It holds S(w,0) = 0 for all w ∈ H 1 (S 1 ). Moreover, the functional S is continuous and compact (since the embedding H 2 (S 1 ) ⊂ H 1 (S 1 ) is compact). We need to prove a uniform bound for all fixed points of S(·,σ). Let y be a fixed point of S(·,σ), i.e., y ∈ H 2 (S 1 ) solves for all φ ∈ H 2 (S 1 ) Z Z σ (ez − ey )φdx. (2.5) (ey yxx φxx + εyx φx + εyφ)dx = τ 1 1 S S Using the test function φ = 1 − e−y yields Z Z Z Z 2 yxx dx − yxx yx2 dx + ε e−y yx2 dx + ε y(1 − e−y )dx S1 S1 S1 S1 Z σ (ez − ey )(1 − e−y )dx. = τ S1 The second term on the left-hand side vanishes since yxx yx2 = (yx3 )x /3. The third and fourth term on the left-hand side are non-negative. Furthermore, with the inequality ex ≥ 1 + x for all x ∈ R, (ez − ey )(1 − e−y ) ≤ (ez − z) − (ey − y). We obtain σ τ

Z

Z (ey − y)dx +

S1

S1

2 yxx dx ≤

σ τ

Z (ez − z)dx. S1

As z is given, this provides a uniform bound for yxx in L2 (S 1 ). Moreover, the inequality ex − x ≥ |x| for all x ∈ R implies a (uniform) bound for y in L1 (S 1 ) and for R ydx. Now we use the Poincaré inequality µ ¶2 Z ° dx ° L L ° ° u ° ≤ kux kL2 (S 1 ) ≤ kuxx kL2 (S 1 ) for all u ∈ H 2 (S 1 ). °u − 2 1 L L (S ) 2π 2π S1 R Recall that kuk2L2 (S 1 ) = S 1 u2 dx/L. Then the above estimates provide a (uniform in ε) bound for y and yx in L2 (S 1 ) and thus for y in H 2 (S 1 ). This shows that all fixed points of the operator S(·,σ) are uniformly bounded in H 1 (S 1 ). We notice that we even obtain a uniform bound for y in H 2 (S 1 ) which is independent of ε. The LeraySchauder fixed-point theorem finally ensures the existence of a fixed point of S(·,1), i.e., of a solution y ∈ H 2 (S 1 ) to (2.3). Next we show that the limit ε → 0 can be performed in (2.3) and that the limit function satisfies (2.2). Let yε be a solution of (2.3). The above estimate shows that yε is bounded in H 2 (S 1 ) uniformly in ε. Thus there exists a subsequence (not relabeled) such that, as ε → 0, yε * y yε → y

weakly in H 2 (S 1 ), strongly in H 1 (S 1 ) and in L∞ (S 1 ).

We conclude that eyε → ey in L2 (S 1 ) as ε → 0. In particular, eyε (yε )xx * ey yxx weakly in L1 (S 1 ). The limit ε → 0 in (2.5) can be performed proving that y solves (2.2).

280


Moreover, using R R the test function R φ ≡ 1 in the weak formulation of (2.2) shows that exp(yk )dx = exp(yk−1 )dx = u0 dx. It remains to prove that the solution yk of (2.2) lies in C ∞ (S 1 ). Writing u = eyk , we can reformulate (2.2) as µ 2 ¶ 1 2ux uxx u3x uxxxx = − 2 − (u − eyk−1 ). (2.6) u u x τ Since yk ∈ H 2 (S 1 ) ,→ W 1,∞ (S 1 ), also u ∈ H 2 (S 1 ) ,→ W 1,∞ (S 1 ) and u is strictly positive in S 1 . In particular, 1/u ∈ L∞ (S 1 ). Thus u2x uxx /u ∈ L2 (S 1 ) and u3x /u2 ∈ L∞ (S 1 ). By (2.6) this implies that uxxxx ∈ H −1 (S 1 ) and therefore u ∈ H 3 (S 1 ). Again by (2.6) this yields uxxxx ∈ L2 (S 1 ) and u ∈ H 4 (S 1 ). Continuing this procedure leads to u ∈ H n (S 1 ) for all n ∈ N and hence u ∈ C ∞ (S 1 ). Finally, since u is strictly positive, this shows that yk = logu ∈ C ∞ (S 1 ). For the proof of Theorem 2.1 we need further uniform estimates for the finite sequence (y (N ) ). For this, let y (N ) be defined by y (N ) (x,t) = yk (x) for x ∈ S 1 , t ∈ (tk−1 ,tk ], 1 ≤ k ≤ N . Then we have shown in the proof of Lemma 2.2 that there exists a constant c > 0 depending neither on τ nor on N such that ky (N ) kL2 (0,T ;H 2 (S 1 )) + ky (N ) kL∞ (0,T ;L1 (S 1 )) + key

(N )

kL∞ (0,T ;L1 (S 1 )) ≤ c.

(2.7)

To pass to the limit in the approximating equation, we need further compactness (N ) estimates on ey . Here we proceed similarly as in [11]. Lemma 2.3. The following estimates hold: ky (N ) kL5/2 (0,T ;W 1,∞ (S 1 )) + key

(N )

kL5/2 (0,T ;W 1,1 (S 1 )) ≤ c,

(2.8)

where c > 0 does not depend on τ and N . Proof. We obtain from the Gagliardo-Nirenberg inequality and (2.7): 3/5

2/5

1/5

4/5

ky (N ) kL5/2 (0,T ;L∞ (S 1 )) ≤ cky (N ) kL∞ (0,T ;L1 (S 1 )) ky (N ) kL1 (0,T ;H 2 (S 1 )) ≤ c, kyx(N ) kL5/2 (0,T ;L∞ (S 1 )) ≤ cky (N ) kL∞ (0,T ;L1 (S 1 )) ky (N ) kL2 (0,T ;H 2 (S 1 )) ≤ c. This implies the first bound in (2.8). The second bound follows from the first one and (2.7): (N )

key kL5/2 (0,T ;W 1,1 (S 1 )) ³ (N ) ´ (N ) ≤c key kL5/2 (0,T ;L1 (S 1 )) + k(ey )x kL5/2 (0,T ;L1 (S 1 )) ≤ckey ≤c.

(N )

kL5/2 (0,T ;L1 (S 1 )) + ckey

(N )

kL∞ (0,T ;L1 (S 1 )) kyx(N ) kL5/2 (0,T ;L∞ (S 1 ))

The lemma is proved. We also need an estimate for the discrete time derivative. We introduce the shift operator σN by (σN (y (N ) ))(x,t) = yk−1 (x) for x ∈ S 1 , t ∈ (tk−1 ,tk ]. Lemma 2.4. The following estimate holds: key

(N )

− eσN (y

(N )

)

kL10/9 (0,T ;H −2 (0,1)) ≤ cτ,

(2.9)


281

where c > 0 does not depend on τ and N . Proof. From (2.2) and Hölder’s inequality we obtain (N ) (N ) 1 y(N ) (N ) ke − eσN (y ) kL10/9 (0,T ;H −2 (S 1 )) ≤ key yxx kL10/9 (0,T ;L2 (S 1 )) τ (N ) (N ) ≤ key kL5/2 (0,T ;L∞ (S 1 )) kyxx kL2 (0,T ;L2 (S 1 )) ,

and the right-hand side is uniformly bounded by (2.7) and (2.8) since W 1,1 (0,1) ,→ L∞ (0,1). Now we are able to prove Theorem 2.1, i.e. to perform the limit τ → 0 in (2.2). From estimate (2.7) the existence of a subsequence of y (N ) (not relabeled) follows such that, as N → ∞ or, equivalently, τ → 0, y (N ) * y

weakly in L2 (0,T ;H 2 (S 1 )).

(2.10)

Since the embedding W 1,1 (S 1 ) ⊂ L2 (S 1 ) is compact it follows from the second bound in (2.8) and from (2.9) by an application of Aubin’s lemma [17, Thm. 5] that, up to (N ) the extraction of a subsequence, ey → g strongly in L5/2 (0,T ;L2 (S 1 )). (N ) We claim that g = ey . For this, let z be a smooth function. Since ey → g strongly in L2 (0,T ;L2 (S 1 )) and y (N ) * y weakly in L2 (0,T ;L2 (S 1 )), we can pass to the limit N → ∞ in Z TZ (N ) 0≤ (ey − ez )(y (N ) − z)dxdt S1

0

to obtain the inequality Z

T

Z (g − ez )(y − z)dxdt.

0≤ 0

S1

The monotonicity of x 7→ ex finally yields g = ey . Noticing that the uniform estimate (2.9) implies, for a subsequence, ´ (N ) 1 ³ y(N ) e − eσN (y ) * (ey )t weakly in L10/9 (0,T ;H −2 (S 1 )), τ

(2.11)

we can pass to the limit τ → 0 in (2.2), using the convergence results (2.10)-(2.11), which concludes the proof of Theorem 2.1. 3. Regularity of solutions Theorem 3.1. Let u0 ∈ H 1 (S 1 ) satisfy the assumptions of Theorem 2.1. Then the solution constructed in Theorem 2.1 satisfies the regularity properties √ u ∈ L2 (0,T ;H 3 (S 1 )) ∩ L∞ (0,T ;H 1 (S 1 )). (3.1)

The theorem is an immediate consequence of the following lemma and the convergence properties shown in the previous section. Lemma 3.2. Let yk ∈ C ∞ (S 1 ) be the solution of (2.2) constructed in Theorem 2.1 and set uk = eyk , k = 1,...,N . Then Z Z ¢ ¡ √ √ 1 √ |( uk )xxx |2 dx ≤ 0, |( uk )x |2 − |( uk−1 )x |2 dx + µ τ S1 S1

282


√ where µ = (103 + 241)/72 = 1.646169... is the largest root of 36x2 − 103x + 72 = 0. Proof. The proof is based on the algorithmic entropy construction method recently developed in [12]. The main idea of this method is to reformulate the task of proving entropy dissipation employing integration by parts as a decision problem for polynomial systems. The functions uk = eyk and uk−1 = eyk−1 are strictly positive and smooth and satisfy 1 (uk − uk−1 ) + (uk (loguk )xx )xx = 0 in S 1 . τ −1/2

(3.2)

1/2

We multiply this equation by uk (uk )xx , integrate over S 1 , and integrate by parts the second term: Z 1 −1/2 1/2 0= u (uk − uk−1 )(uk )xx dx τ S1 k Z ¡ ¢ ¡ −1/2 1/2 ¢ − uk (loguk )xx x uk (uk )xx x dx =: I1 − I2 . S1

Integrating by parts once, an elementary computation shows that Z Z ¡ 1/2 2 ¢ 1 1 1/2 I1 = − |(uk )x | − |(uk−1 )x |2 dx − uk−1 |(loguk − loguk−1 )x |2 dx τ S1 4τ S 1 Z ¡ √ ¢ 1 √ |( uk )x |2 − |( uk−1 )x |2 dx. ≤− τ S1 We claim that the integral I2 can be estimated as Z √ I2 ≥ µ |( uk )xxx |2 dx,

(3.3)

S1

where µ is as in the statement of the lemma. We write, omitting the index k in the following, Z ³ u ´3 u ³ u ´2 ³ u ´2 h 1 ³ u ´6 ³ u ´4 u x x x x xx xxx xx + +2 I2 = u −2 2 u u u u u u u 1 S ux uxx uxxx 1 ³ uxxx ´2 i + dx. −2 u u u 2 u Moreover, we have Z Z √ 2 J= ( u)xxx dx =

h 9 ³ u ´6 9 ³ u ´4 u 3 ³ ux ´3 uxxx x x xx − + 64 u 16 u u 8 u u S1 S1 ³ ´ ³ ´ ³ ´ i 2 2 2 9 ux uxx 3 ux uxx uxxx 1 uxxx + − + dx. 16 u u 4 u u u 4 u u

In order to show that inequality (3.3) holds, we need to perform suitable integrations by parts. It turns out that only the following integration by parts is useful: Z ³ 5´ Z h ³ u ´6 ³ u ´4 u i ux x x xx 0= dx = dx = J1 , u − 4 + 5 4 x u u u S1 u S1 Z ³ 3 Z h ³ u ´4 u ³ u ´3 u ³ u ´2 ³ u ´2 i ux uxx ´ xx xxx xx x x x dx = J2 . 0= dx = u − 3 + + 3 u3 u u u u u u x S1 S1


283

Then we can write the integral as I2 = I2 + c1 J1 + c2 J2 for arbitrary constants c1 , c2 ∈ R. We wish to find µ > 0 and c1 , c2 ∈ R such that I2 + c1 J1 + c2 J2 ≥ µJ. For this task we identify the derivative ∂xj u/u with the real variable ξj and deal with the polynomials 1 1 P (ξ) = ξ16 − 2ξ14 ξ2 + ξ13 ξ3 + 2ξ12 ξ22 − 2ξ1 ξ2 ξ3 + ξ32 , which corresponds to I2 , 2 2 9 6 9 4 3 3 9 2 2 3 1 E(ξ) = ξ1 − ξ1 ξ2 + ξ1 ξ3 + ξ1 ξ2 − ξ1 ξ2 ξ3 + ξ32 , which corresponds to J, 64 16 8 16 4 4 T1 (ξ) = −4ξ16 + 5ξ14 ξ2 , which corresponds to J1 , T2 (ξ) = −3ξ14 ξ2 + ξ13 ξ3 + 3ξ12 ξ22 , which corresponds to J2 . Thus we wish to find constants µ > 0 and c1 , c2 ∈ R such that (P − µE + c1 T1 + c2 T2 )(ξ) ≥ 0 for all ξ = (ξ1 ,ξ2 ,ξ3 )> ∈ R3 , which corresponds to a pointwise estimate of the integrand of I2 . The determination of all parameters such that the above inequality is true is called a quantifier elimination problem. In this situation it can be explicitly solved. The above inequality is equivalent to a1 ξ16 + a2 ξ14 ξ2 + a3 ξ13 ξ3 + a4 ξ12 ξ22 + a5 ξ1 ξ2 ξ3 + a6 ξ32 ≥ 0

for all ξ1 ,ξ2 ,ξ3 ∈ R,

where 1 9 a1 = − µ − 4c1 , 2 64 9 a4 = 2 − µ + 3c2 , 16

9 µ + 5c1 − 3c2 , 16 3 a5 = −2 + µ, 4

a2 = −2 +

3 a3 = 1 − µ + c2 , 8 1 1 a6 = − µ. 2 4

(3.4) (3.5)

Now, we recall (a slight generalization of) Lemma 12 of [12]: Lemma 3.3. [12] Let the real polynomial P (ξ1 ,ξ2 ,ξ3 ) = a1 ξ16 + a2 ξ14 ξ2 + a3 ξ13 ξ3 + a4 ξ12 ξ22 + a5 ξ1 ξ2 ξ3 + a6 ξ32 be given and let a6 > 0. Then the statement P (ξ1 ,ξ2 ,ξ3 ) ≥ 0

for all ξ1 ,ξ2 ,ξ3 ∈ R

is equivalent to either or

(i)

4a4 a6 − a25 = 2a2 a6 − a3 a5 = 0

(ii)

4a4 a6 − a25 > 0

and

4a1 a6 − a23 ≥ 0,

(3.6)

4a1 a4 a6 − a1 a25 − a22 a6 − a23 a4 + a2 a3 a5 ≥ 0.

(3.7)

and

In both cases, the condition a6 > 0, which is equivalent to µ < 2, has to be satisfied. The first two conditions of case (i) give c2 =

µ , 24(2 − µ)

c1 =

µ(3µ − 8) , 240(2 − µ)2

and the third one is equivalent to −36µ2 + 103µ − 72 ≥ 0.

284


This polynomial has two real roots, √ 103 − 241 = 1.214942... 72

and

√ 103 + 241 = 1.646169... 72

√ Thus we obtain the requirement µ ≤ (103 + 241)/72 in case (i). It can be seen that case (ii) gives a stronger condition on µ; we leave the details to the reader. Remark 3.4. From Theorem 3.1 it follows that u ∈ L∞ (S 1 × (0,∞)) ∩ H 1 (0,T ;H −1 (S 1 ))

for all T > 0.

Indeed, we know that u(N ) ∈ L∞ (0,T ;H 1 (S 1 )) ∩ L2 (0,T ; H 3 (S 1 )). Thus, the first property is a consequence of the embedding H 1 (S 1 ) ,→ L∞ (S 1 ) in one space dimension. Furthermore, we obtain from (3.2) p ¢ ¡p 1 ¡ (N ) u − σN (u(N ) ) = − 2 u(N ) ( u(N ) )xxx τ p p

−( u(N ) )x ( u(N ) )xx

¢ x

∈ L2 (0,T ;H −1 (S 1 )).

Then the limit N → ∞ shows the claim. 4. Optimal logarithmic Sobolev inequality on S 1 The main goal of this section is the proof of a logarithmic Sobolev inequality for periodic functions. The following theorem is due to Weissler and Rothaus (see [9, 16, 18]). We give a simple proof using the entropy–entropy production method. Recall that S 1 is parametrized by 0 ≤ x ≤ L. R Theorem 4.1. Let H1 = {u ∈ H 1 (S 1 ) : ux 6≡ 0 a.e.} and kuk2L2 (S 1 ) = S 1 u2 dx/L. Then R

inf R

u∈H1

u2 dx S1 x u2 log(u2 /kuk2L2 (S 1 ) )dx S1

=

2π 2 . L2

(4.1)

We recall that the optimal constant in the usual Poincaré inequality is L/2π, i.e. R inf R

v∈H1

where v¯ =

R S1

S1

v 2 dx S1 x (v − v¯)2 dx

=

4π 2 , L2

(4.2)

vdx/L.

Proof. Let I denote the value of the infimum in (4.1). First we prove that R 2 1 1 v dx I ≤ inf R S x 2 , 2 v∈H1 S 1 (v − v¯) dx

(4.3)

since then the upper bound I ≤ 2π 2 /L2 follows directly from (4.2). Let v ∈ H1 and set u R = uε = 1 + ε(v − v¯).2 Without loss2 of2 generality, we may replace v − v¯ by 2v such that vdx = 0. Then u = 1 + 2εv + ε v and the expansion log(1 + x) = x − x /2 + O(x3 ) S1


285

for x → 0 yields Z

Z u2 log(u2 )dx =

(1 + 2εv + ε2 v 2 )log(1 + 2εv + ε2 v 2 )dx Z = 3ε2 v 2 dx + O(ε3 ), S1 ¶ Z µ Z ¶ µ Z Z 1 1 2 2 2 2 2 2 u dx = (1 + ε v )dxlog (1 + ε v )dx u dxlog L S1 L S1 S1 S1 Z = ε2 v 2 dx + O(ε4 ) (ε → 0). S1

S1

S1

Taking the difference of the two expansions gives µ ¶ Z Z u2 2 2 u log R dx = 2ε v 2 dx + O(ε3 ). 2 dx/L u 1 1 S1 S S R R 2 2 Therefore, using S 1 ux dx = ε S 1 vx2 dx, R R u2 dx 1 S 1 vx2 dx S1 x R R = + O(ε). u2 log(u2 /kuk2L2 (S 1 ) )dx 2 S 1 v 2 dx S1 In the limit ε → 0 we obtain (4.3). In order to prove the lower bound for the infimum we use the entropy–entropy production method. For this we consider the heat equation vt = vxx

in S 1 × (0,∞),

v(·,0) = u2

in S 1

R for some function u ∈ H 1 (S 1 ). We assume for simplicity that kuk2L2 (S 1 ) = S 1 u2 dx/L = 1. Then Z Z d v logvdx = −4 wx2 dx, dt S 1 1 S √ where the function w := v solves the equation wt = wxx + wx2 /w. Now, the time derivative of Z Z 2π 2 f (t) = wx2 dx − 2 w2 log(w2 )dx L 1 1 S S equals Z 0

f (t) = −2 S1

µ 2 wxx +

¶ Z wx4 4π 2 2 2 wx4 − w dx ≤ − dx ≤ 0, x 3w2 L2 3 S 1 w2

where we have used the Poincaré inequality Z Z L2 w2 dx. wx2 dx ≤ 2 4π S 1 xx S1 This shows that f (t) is non-increasing and moreover, for any u ∈ H 1 (S 1 ), ! Ã Z Z 2π 2 u2 2 2 dx = f (0) ≥ f (t). ux dx − 2 u log L S1 kuk2L2 (S 1 ) S1

(4.4)

286


As the solution v(·,t) of the above heat equation and hence w(·,t) converges to zero in appropriate Sobolev norms as t → +∞, we conclude that f (t) → 0 as t → +∞. This implies I ≥ 2π 2 /L2 . Remark 4.2. Similar results as in Theorem 4.1 can be obtainedRfor the so-called convex Sobolev inequalities. Let σ(v) = (v p − v¯p )/(p − 1), where v¯ = S 1 vdx/L for 1 < p ≤ 2. We claim that R 00 2 8π 2 1 σ (v)vx dx inf SR = 2. v∈H1 L σ(v)dx S1 As in the logarithmic case, the lower bound is achieved by an expansion around 1 and the usual Poincaré inequality. On the other hand, let v be a solution of the heat equation. Then Z Z d 4 σ(v)dx = − w2 dx dt S 1 p S1 x where w = v p/2 solves µ wt = wxx +

¶ 2 wx 2 −1 , p w

(4.5)

and, using (4.4), ¶ µ ¶ 4 ¶ Z µ Z µ d 2π 2 p 4π 2 2 2 wx 2 2 wx − 2 σ(v) dx = −2 wxx − 2 wx + −1 dx 2 dt S 1 L L p 3w 1 S µ ¶Z wx4 2 2 −1 dx ≤ 0. ≤− 2 3 p S1 w This proves the upper bound Z Z Z p 2π 2 p σ 00 (v)vx2 dx = wx2 dx ≥ 2 σ(v)dx. 4 S1 L S1 S1 With the notation v = u2/p this result takes the more familiar form ·Z µ Z ¶p ¸ Z 1 1 L2 u2 dx − L u2/p dx ≤ 2 u2 dx for all u ∈ H 1 (S 1 ). p − 1 S1 L S1 2π p S 1 x

(4.6)

The logarithmic case corresponds to the limit p → 1 whereas the case p = 2 gives the usual Poincaré inequality. We may notice that the method gives more than what is stated in Theorem 4.1 since there is an integral remainder term. Namely, for any p ∈ [1,2], for any v ∈ H 1 (S 1 ), we have Z Z p 2π 2 p 00 2 σ(v)dx σ (v)vx dx + R[v] ≥ 2 4 S1 L S1 with Z

∞Z

R[v] = 2 0

S1

µ 2 wxx −

µ ¶ 4 ¶ 4π 2 2 2 wx wx + dxdt, −1 2 L p 3w2

287

JEAN DOLBEAULT, IVAN GENTIL AND ANSGAR JUNGEL p/2

where w = w(x,t) is the solution to (4.5) with initial datum u0 . Inequality (4.6) can also be improved with an integral remainder term for any p ∈ [1,2], where in the limit case p = 1, one has to take σ(v) = v log(v/¯ v ). As a consequence, the only optimal functions in (4.1) or in (4.6) are the constants. Corollary 4.3. Let n ∈ N, n > 0 and let Hn = {u ∈ H n (S 1 ) : ux 6≡ 0 a.e.}. Then R S1

inf R

u∈Hn

S1

¯ (n) ¯2 ¯u ¯ dx

u2 log(u2 /kuk2L2 (S 1 ) )dx

=

1 2

µ

2π L

¶2n .

(4.7)

Proof. We obtain a lower bound by applying successively Theorem 4.1 and the Poincaré inequality: Ã ! µ ¶2n Z ¯ Z Z ¯ L2 L u2 ¯ (n) ¯2 2 2 dx ≤ u dx ≤ 2 u log ¯u ¯ dx. kuk2L2 (S 1 ) 2π 2 S 1 x 2π S1 S1 The upper bound is achieved as in the proof of Theorem 4.1 by expanding the R quotient for u = 1 + εv with S 1 vdx = 0 in powers of ε, R

R

¯ (n) ¯2 ¯u ¯ dx

1 = 2 log(u2 /kuk2 2 u )dx L2 (S 1 ) S1 S1

R 1 SR

¯ (n) ¯2 ¯v ¯ dx S1

v 2 dx

+ O(ε)

(ε → 0),

and using the Poincaré inequality R

¯ (n) ¯2 ¯v ¯ dx

µ

2π inf R = 2 u∈Hn L |v − v¯| dx S1 S1

¶2n .

2n

The best constant ω = (2π/L) in such an inequality is easily recovered by looking for the smallest positive value of ω for which there exists a nontrivial periodic solution of (−1)n v (2n) + ωv = 0. 5. Exponential time decay of the solutions We show the exponential time decay of the solutions of (1.1). Our main result is contained in the following theorem. Theorem 5.1. Assume R R that u0 is a nonnegative measurable function such that (u − logu )dx and u logu0 dx are 0 0 1 S S1 0 R finite. Let u be the weak solution of (1.1) constructed in Theorem 2.1 and set u ¯ = S 1 u0 (x)dx/L. Then µ

¶ Z ³u ´ u(·,t) 0 −M1 t u(·,t)log dx, dx ≤ e u0 log u ¯ u ¯ 1 1 S S

Z

Moreover, if in addition Z S1

√

where M1 =

32π 4 . L4

u0 ∈ H 1 (S 1 ), then

¡√ ¢2 ( u)x (·,t) dx ≤ e−M2 t

Z S1

√ ( u0 )2x dx,

where M2 =

16µπ 4 L4

√ where µ = (103 + 214)/72 = 1.646169... is the largest root of 36x2 − 103x + 72 = 0.

288


Proof. Since we do not have enough regularity of the solutions of (1.1) in order to manipulate the differential equation directly, we need to regularize the equation first. For this we consider the semi-discrete problem 1 (uk − uk−1 ) + (uk (loguk )xx )xx = 0 τ

in S 1

(5.1)

as in the proof of Theorem 2.1. The solution uk ∈ C ∞ (S 1 ) of this problem for given uk−1 is strictly positive and we can use loguk as a test function in the weak formulation of (5.1). In order to simplify the presentation we set u := uk and z := uk−1 . Then we obtain as in [15] Z Z 1 2 (ulogu − z logz)dx + u|(logu)xx | dx ≤ 0. (5.2) τ S1 1 S From integration by parts it follows Z Z u2x uxx 2 u4x dx = dx. 2 u 3 S 1 u3 S1 This identity gives ¶ ¶ Z Z µ 2 Z µ 2 uxx u4x uxx u2x uxx 1 u4x 2 u|(logu)xx | dx = + 3 −2 2 dx = − dx u u u u 3 u3 S1 S1 S1 ¶ Z µ 2 Z Z √ uxx 1 u4x uxx u2x 1 u4x 2 = u) | dx + + − dx = 4 |( dx. xx u 3 u3 u2 12 S 1 u3 S1 S1 Thus, (5.2) becomes Z ³ Z ³u´ ³ z ´´ ¯√ ¯ 1 ¯( u)xx ¯2 dx ≤ 0. ulog − z log dx + 4 τ S1 u ¯ u ¯ S1

(5.3)

Now we use Corollary 4.3 with n = 2: Z Z ³u´ ¯√ ¯ L4 ¯( u)xx ¯2 dx. dx ≤ 4 ulog u ¯ 8π S 1 S1 From this inequality and (5.3) we conclude Z ³ Z ³u´ ³u´ ³ z ´´ 1 32π 4 ulog ulog − z log dx + 4 dx ≤ 0. τ S1 u ¯ u ¯ L u ¯ S1 This is a difference inequality for the sequence Z ³u ´ k dx, Ek := uk log u ¯ S1 yielding (1 + τ M1 )Ek ≤ Ek−1

or Ek ≤ E0 (1 + τ M1 )−k ,

where M1 is as in the statement of the theorem. For t ∈ ((k − 1)τ,kτ ] we obtain further Ek ≤ E0 (1 + τ M1 )−t/τ .


289

Now the proof goes exactly as in [15]. Indeed, the functions uk (x) converge a.e. to u(x,t) and (1 + τ M1 )−t/τ → e−M1 t as τ → 0. This implies the first assertion. The second one is obtained similarly employing Lemma 3.2. √ Remark 5.2. In the H 1 -norm we obtain the following decay estimate if u0 ∈ H 1 (S 1 ): ku(·,t) − u ¯k2H 1 (S 1 ) ≤ Ce−M2 t ,

(5.4)

where 16µπ 4 M2 = L4

³

L2 ´ and C = 4 1 + 2 4π

"√

√ L √ ¯ k( u0 )x kL2 (S 1 ) + u 2

#2

√ k( u0 )x k2L2 (S 1 ) .

The decay rate is slightly worse than that in [5] which equals 32π 4 /L4 , but we do not need the strong condition k(logu0 )x kL2 (S 1 ) < 12 which is assumed in [5]. The proof of (5.4) uses the inequality √ L kg − g¯kL∞ (S 1 ) ≤ kgx kL2 (S 1 ) for all g ∈ H 1 (S 1 ) 2 and the uniform bound p

√

√

√

√ L √ ¯kL∞ (S 1 ) + u ¯≤ ¯ ku(·,t)kL∞ (S 1 ) ≤ k u(·,t) − u k( u)x (·,t)kL2 (S 1 ) + u 2 √ √ L √ ≤ k( u0 )x kL2 (S 1 ) + u ¯. 2 By Poincaré’s inequality, we have Z ³ ¡ √ √ ¢2 L2 ´ 2 u( u)x (·,t)dx ku(·,t) − u ¯k2H 1 (S 1 ) ≤ 1 + 2 4π S1 ³ √ L2 ´ p ≤ 4 1 + 2 k u(·,t)k2L∞ (S 1 ) k( u)x (·,t)k2L2 (S 1 ) 4π √ ³ √ ´2 √ L2 ´³ L √ ≤4 1+ 2 k( u0 )x kL2 (S 1 ) + u ¯ k( u0 )x k2L2 (S 1 ) e−M2 t . 4π 2

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